Saturday, February 27, 2016

The original neon-bulb oscillator circuit studied by Van der Pol and Van der Mark in their 1927 paper consists of a neon-bulb connected to a capacitor, resistor, a bias voltage source and a sinuosoidal signal voltage source. The operation of this circuit without the sinusoidal signal can be described as follows: the neon bulb is not conducting and the voltage source charges up the the capacitor until a certain point at which the neon bulb starts conducting (and glowing) and discharges the capacitor until the voltage is low enough to turn off the neon bulb and the cycle starts anew. Looking at the $v$-$i$ characteristic of the neon bulb we see that it is a function with two local extrema just like the cubic polynomial. The role of the bias voltage source is to add power to the system, since the neon bulb by itself does not supply any power. In particular, by looking at the equivalent $v$-$i$ characteristic of the neon bulb in parallel with the bias voltage, it effectively shift the characteristic so it looks more like the cubic and thus can oscillate.

Except for the external driving source, this circuit is similar to the circuits considered in a previous post. To write state equations, we model the neon-bulb as a nonlinear resistor in parallel with an small linear inductor and a small linear capacitor.

When the sinusoidal source is turned off, the oscillator oscillates freely with a frequency $\omega_1$. The addition of the sinusoidal voltage source add another frequency $\omega_2$ to the system. One can think of the Poincaré map (generated by sampling the flow at frequency $\omega_2$) as a map on the circle, the circle being the limit cycle. In fact, for certain parameters, we see behavior similar to those obtained from the circle map.

By adjusting the capacitor $C$, we change the free running frequency of the oscillator, i.e. changing the ratio between the two frequency. We observe frequency locking where the oscillator oscillates at a submultiple of the driving frequency for a range of capacitor values. If we vary the driving frequency, we observe the Devil's staircase, similar to the one observed in the circle map.

A triangular number $T(n)$ is defined as $n(n+1)/2$. It is equal to the binomial coefficient $\left(\begin{array}{c}n+1\\2\end{array}\right)$ and is also equal to $0+1+2+\cdots+n$. When is an integer $m\geq 0$ equal to a triangular number $T(n)$ for some integer $n\geq 0$? We have $n(n+1) = 2m$. Using the quadratic formula to solve the quadratic equation $n^2+n-2m = 0$ for $n$, we obtain one solution for $n$:
$$ n = \frac{\sqrt{8m+1}-1}{2} $$
The other solution for $n$ is negative (or complex) so we will not use that.
If $8m+1$ is not the square of an integer, then $n$ is not an integer. If $8m+1$ is the square of an integer, then $8m+1$ is odd and thus $\sqrt{8m+1}$ is odd as well. This means that $n = \frac{\sqrt{8m+1}-1}{2}$ is an integer. Thus $m$ is a triangular number if and only if $8m+1$ is the square of an integer.

Sunday, February 14, 2016

We were driving on the road today and listening to "In the shadow of the mountain king" from Peer Gynt by Edvard Grieg when my son remarked: "This is not a good piece of music to listen to when driving, it makes you drive faster and faster." My wife replied: "The same is true for Bolero". I tend to agree. Luckily, my driving is not affected by the tempo of the music and we reached our destination safely.

Friday, February 12, 2016

Exponential growth is a common phenomena occurring all around us. Sometimes it is good, other times not. Simply speaking, we talk about exponential growth if some quantity is increasing as an exponential function of another quantity, i.e. y = abx.

The Good:

Typically when you put your money in the bank, your banks pays you interest as a percentage of your money, i.e. if you have \$100 in the bank and the bank pays you 5% interest a year, then you will have 1.05$\times$\$100 = \$105 after one year. Assuming the interest rate is constant, if the next year, the bank only pays you 5% of the original \$100, then your gain will be \$5 year after year, i.e the money in the bank is 100+5n after n years. But because of compound interest, i.e. your interest will also earn interest, you will receive 5% of \$105 = \$5.25 in interest the second year and the money in the bank will be will be \$100*1.05n after n years, i.e. an exponential growth. If you can appreciate how fast exponential growth is, this is the best reason for anyone to save money early.

In cryptography, making the number of possibilities that an encryption key can take on large is an important idea to make cryptographic techniques secure. This is akin to choosing a long password. The number of combinations grows exponentially with the number of digits. For instance, a n-digit decimal number has $10^n$ possible combinations.

The not so Good:

In the analysis of computer algorithms, many known algorithms for solving hard NP-complete problems (such as 3SAT) have time complexity exponential in the problem size. This means that the size of the problem that can be solved will be limited.

Many models of growth assumes an exponential growth rate, but this cannot be sustained indefinitely in a finite universe. Some examples are Moore's Law or models of population growth. Models of chaotic systems such as the Lorenz system and Chua's circuit also experience exponential divergence of trajectories on a small scale, but on a larger scale the trajectories remain bounded.

Monday, February 8, 2016

Consider the binomial coefficient C(n,m) = n*(n-1)*...*(n-m+1)/(m*(m-1)*...*2*1). When is this number odd and when is it even? In this post, we use Lucas' theorem to provides a simple characterization of this. Lucas' theorem states that for integers $m$, $n$ and prime $p$, the following relationship holds:
\[ C(n,m) \equiv \prod_{i=0}^{k} C(n_i, m_i) \mod p \]
where $n_i$ and $m_i$ are the digits of $n$ and $m$ in base $p$ respectively, with $n_0$ and $m_0$ being the least significant digits.

When $p=2$, $n_i$ and $m_i$ are the bits in the binary expansion of $n$ and $m$ and $C(n_i, m_i)$ is $0$ if and only if $n_i < m_i$. This implies that $C(n,m) \equiv 0 \mod 2$ if and only if $n_i < m_i$ for some $i$.

Incidentally, for bits $n_i$ and $m_i$, $n_i < m_i$ is logically equivalent to $NOT (m_i\Rightarrow n_i)$.
Next we consider C(n,m)C(a,b) mod 2. Clearly this is equivalent (mod 2) to (C(n,m) mod 2)*(C(a,b) mod 2). Thus $C(n,m)C(a,b) \equiv 0 mod 2$ if and only if $m AND ( NOT n)$ is not zero or $b AND ( NOT a)$ is not zero. This in turns implies the following: